💡Alright, so I have this problem about a shopping mall estimating the demand and profit for a certain product over the months of 2018. Let me try to understand what's being asked here.First, the demand function is given as ( f(x) = -3x^2 + 40x ), where ( x ) is the month number, ranging from 1 to 12. So, this is a quadratic function that models how the demand changes each month. Since the coefficient of ( x^2 ) is negative, I know this is a downward-opening parabola, meaning the demand will increase to a certain point and then decrease.Next, the monthly purchase unit price is given by ( q(x) = 150 + 2x ). This is a linear function, so the purchase price increases by 2 Yuan each month. The selling price is fixed at 185 Yuan per item. So, the profit per item sold would be the selling price minus the purchase price, which is ( 185 - q(x) ).To find the monthly profit, I need to multiply the profit per item by the number of items sold, which is the demand ( f(x) ). So, the profit function ( y ) should be:[y = [185 - q(x)] times f(x)]Substituting the given functions:[y = [185 - (150 + 2x)] times (-3x^2 + 40x)]Simplifying the expression inside the brackets first:[185 - 150 - 2x = 35 - 2x]So, the profit function becomes:[y = (35 - 2x)(-3x^2 + 40x)]Now, I need to multiply these two expressions together. Let's do that step by step.First, multiply 35 by each term in the second polynomial:[35 times (-3x^2) = -105x^2][35 times 40x = 1400x]Next, multiply -2x by each term in the second polynomial:[-2x times (-3x^2) = 6x^3][-2x times 40x = -80x^2]Now, combine all these terms:[y = 6x^3 - 80x^2 - 105x^2 + 1400x]Combine like terms:[y = 6x^3 - (80x^2 + 105x^2) + 1400x][y = 6x^3 - 185x^2 + 1400x]So, the profit function is:[y = 6x^3 - 185x^2 + 1400x]Now, to find the maximum profit, I need to find the value of ( x ) that maximizes this function. Since this is a cubic function, it can have one or two critical points. To find the critical points, I need to take the derivative of ( y ) with respect to ( x ) and set it equal to zero.Let's compute the derivative:[y' = frac{d}{dx}(6x^3 - 185x^2 + 1400x) = 18x^2 - 370x + 1400]So, the derivative is:[y' = 18x^2 - 370x + 1400]To find the critical points, set ( y' = 0 ):[18x^2 - 370x + 1400 = 0]This is a quadratic equation in the form ( ax^2 + bx + c = 0 ), where ( a = 18 ), ( b = -370 ), and ( c = 1400 ).I can solve this using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Plugging in the values:[x = frac{-(-370) pm sqrt{(-370)^2 - 4 times 18 times 1400}}{2 times 18}][x = frac{370 pm sqrt{136900 - 100800}}{36}][x = frac{370 pm sqrt{36100}}{36}][x = frac{370 pm 190}{36}]So, we have two solutions:1. ( x = frac{370 + 190}{36} = frac{560}{36} = frac{140}{9} approx 15.56 )2. ( x = frac{370 - 190}{36} = frac{180}{36} = 5 )Since ( x ) must be an integer between 1 and 12, ( x = 15.56 ) is outside the valid range. Therefore, the only critical point within our domain is ( x = 5 ).Now, to determine whether this critical point is a maximum, I can use the second derivative test.First, compute the second derivative:[y'' = frac{d}{dx}(18x^2 - 370x + 1400) = 36x - 370]Evaluate the second derivative at ( x = 5 ):[y''(5) = 36 times 5 - 370 = 180 - 370 = -190]Since ( y''(5) < 0 ), the function is concave down at ( x = 5 ), indicating a local maximum.Therefore, the maximum profit occurs at ( x = 5 ).Now, let's calculate the maximum profit by plugging ( x = 5 ) back into the profit function:[y = 6(5)^3 - 185(5)^2 + 1400(5)][y = 6 times 125 - 185 times 25 + 1400 times 5][y = 750 - 4625 + 7000][y = 750 - 4625 = -3875][y = -3875 + 7000 = 3125]So, the maximum monthly profit is 3125 Yuan, occurring in the 5th month of 2018.Just to be thorough, I should check the profit for the months around ( x = 5 ) to ensure that ( x = 5 ) indeed gives the maximum profit.Let's compute the profit for ( x = 4 ):[y = 6(4)^3 - 185(4)^2 + 1400(4)][y = 6 times 64 - 185 times 16 + 1400 times 4][y = 384 - 2960 + 5600][y = 384 - 2960 = -2576][y = -2576 + 5600 = 3024]Profit for ( x = 4 ) is 3024 Yuan.For ( x = 6 ):[y = 6(6)^3 - 185(6)^2 + 1400(6)][y = 6 times 216 - 185 times 36 + 1400 times 6][y = 1296 - 6660 + 8400][y = 1296 - 6660 = -5364][y = -5364 + 8400 = 3036]Profit for ( x = 6 ) is 3036 Yuan.Comparing these, the profit at ( x = 5 ) is indeed higher than both ( x = 4 ) and ( x = 6 ), confirming that ( x = 5 ) is the month with the maximum profit.Therefore, the shopping mall will have the highest monthly profit for this product in the 5th month of 2018, and the maximum monthly profit is 3125 Yuan.

💡Alright, so I have this problem here about a cubic function and an equation involving its derivative. Let me try to unpack it step by step. First, the function is given as ( f(x) = x^3 + ax^2 + bx + c ). It mentions that this function has extreme points at ( x_1 ) and ( x_2 ). I know that extreme points occur where the first derivative is zero, so I should probably start by finding the derivative of ( f(x) ).Calculating the derivative, I get:[ f'(x) = 3x^2 + 2ax + b ]So, the extreme points ( x_1 ) and ( x_2 ) are the solutions to the equation:[ 3x^2 + 2ax + b = 0 ]That makes sense. Since it's a quadratic equation, there can be at most two real roots, which correspond to the two extreme points mentioned.Next, the problem states that ( f(x_1) = x_1 ). Hmm, that means when I plug ( x_1 ) into the function ( f(x) ), I get ( x_1 ) itself. So, ( x_1 ) is a fixed point of the function ( f(x) ). Interesting. I wonder how this affects the equation we're supposed to solve.The main equation we need to analyze is:[ 3(f(x))^2 + 2af(x) + b = 0 ]I need to find the number of distinct real roots of this equation. Let's denote ( y = f(x) ) for simplicity. Then the equation becomes:[ 3y^2 + 2ay + b = 0 ]So, this is a quadratic equation in terms of ( y ). The solutions to this equation will give me the values of ( y ) (which are ( f(x) )) that satisfy the equation. Once I have those ( y ) values, I can find the corresponding ( x ) values by solving ( f(x) = y ).Let me think about the quadratic equation ( 3y^2 + 2ay + b = 0 ). The discriminant of this quadratic will tell me how many real solutions there are. The discriminant ( D ) is given by:[ D = (2a)^2 - 4 times 3 times b = 4a^2 - 12b ]So, if ( D > 0 ), there are two distinct real roots for ( y ); if ( D = 0 ), there's exactly one real root; and if ( D < 0 ), there are no real roots.But wait, the problem doesn't specify anything about ( a ) and ( b ) other than the fact that ( f(x) ) has extreme points. Since ( f'(x) = 3x^2 + 2ax + b ) has two real roots ( x_1 ) and ( x_2 ), that means the discriminant of ( f'(x) ) must be positive. The discriminant of ( f'(x) ) is:[ D' = (2a)^2 - 4 times 3 times b = 4a^2 - 12b ]Which is the same as the discriminant ( D ) of the quadratic equation in ( y )! So, ( D = D' ). Since ( f'(x) ) has two real roots, ( D' > 0 ), which means ( D > 0 ). Therefore, the quadratic equation ( 3y^2 + 2ay + b = 0 ) has two distinct real roots for ( y ).Let me denote these two roots as ( y_1 ) and ( y_2 ). So, ( y_1 ) and ( y_2 ) are the solutions to ( 3y^2 + 2ay + b = 0 ). Now, for each of these ( y ) values, I need to find the number of real roots ( x ) such that ( f(x) = y ).So, for each ( y_i ) (where ( i = 1, 2 )), the equation ( f(x) = y_i ) is a cubic equation:[ x^3 + ax^2 + bx + c = y_i ]Which simplifies to:[ x^3 + ax^2 + bx + (c - y_i) = 0 ]Since ( f(x) ) is a cubic polynomial, each of these equations can have up to three real roots. However, the actual number of real roots depends on the specific values of ( a ), ( b ), ( c ), and ( y_i ).But wait, there's a condition given in the problem: ( f(x_1) = x_1 ). That means when ( x = x_1 ), ( f(x) = x_1 ). So, ( x_1 ) is a fixed point of the function ( f(x) ). This might imply that ( x_1 ) is a root of the equation ( f(x) - x = 0 ), which is:[ x^3 + ax^2 + bx + c - x = x^3 + ax^2 + (b - 1)x + c = 0 ]But I'm not sure how this directly affects the number of roots for ( f(x) = y_i ). Maybe I need to consider this condition when analyzing the roots.Let me think about the graph of ( f(x) ). Since it's a cubic function, it has an "S" shape, and it tends to negative infinity as ( x ) approaches negative infinity and positive infinity as ( x ) approaches positive infinity. The extreme points ( x_1 ) and ( x_2 ) are where the function changes direction.Given that ( f(x_1) = x_1 ), the point ( (x_1, x_1) ) lies on the graph of ( f(x) ). So, this is a point where the function intersects the line ( y = x ). Similarly, if ( f(x_2) = x_2 ), then ( (x_2, x_2) ) would also lie on the graph, but the problem doesn't specify that. It only mentions ( f(x_1) = x_1 ).Now, going back to the equation ( 3(f(x))^2 + 2af(x) + b = 0 ). We've established that this equation has two distinct real solutions for ( y = f(x) ), namely ( y_1 ) and ( y_2 ). For each of these ( y ) values, we need to find how many real ( x ) satisfy ( f(x) = y ).Since ( f(x) ) is a cubic, each equation ( f(x) = y_i ) can have one or three real roots. The number of real roots depends on the value of ( y_i ) relative to the local maximum and minimum values of ( f(x) ).Given that ( f(x) ) has extreme points at ( x_1 ) and ( x_2 ), the function has a local maximum at one of these points and a local minimum at the other. Let's assume ( x_1 ) is a local maximum and ( x_2 ) is a local minimum, though it could be the other way around depending on the coefficients.Since ( f(x_1) = x_1 ), this means that at the local maximum ( x_1 ), the function value equals ( x_1 ). Similarly, at the local minimum ( x_2 ), the function value is ( f(x_2) ).Now, considering the equation ( f(x) = y_i ), the number of real roots depends on where ( y_i ) is relative to the local maximum and minimum. If ( y_i ) is above the local maximum or below the local minimum, the equation ( f(x) = y_i ) will have only one real root. If ( y_i ) is between the local maximum and minimum, it will have three real roots.But in our case, ( y_1 ) and ( y_2 ) are the solutions to ( 3y^2 + 2ay + b = 0 ). Let me recall that ( y_1 ) and ( y_2 ) are related to the critical points of ( f(x) ). Specifically, since ( f'(x) = 3x^2 + 2ax + b ), and the critical points are at ( x_1 ) and ( x_2 ), the values ( y_1 ) and ( y_2 ) are related to these critical points.Wait a second, if I substitute ( x = x_1 ) into ( f'(x) ), I get zero, which is consistent with it being an extreme point. But how does that relate to ( y_1 ) and ( y_2 )?Let me think differently. The equation ( 3y^2 + 2ay + b = 0 ) can be rewritten as:[ 3y^2 + 2ay + b = 0 ]But notice that ( f'(x) = 3x^2 + 2ax + b ), so if I set ( y = x ), then ( f'(x) = 3x^2 + 2ax + b ). However, in our case, ( y ) is ( f(x) ), not ( x ). So, it's not directly substituting ( x ) into ( f'(x) ).But perhaps there's a relationship here. Let me consider that ( y_1 ) and ( y_2 ) are the values of ( f(x) ) at the critical points. Wait, no, that's not necessarily true. The critical points are where the derivative is zero, but ( y_1 ) and ( y_2 ) are solutions to a different equation.Hmm, maybe I'm overcomplicating this. Let's step back.We have two values ( y_1 ) and ( y_2 ) such that ( 3y_i^2 + 2a y_i + b = 0 ) for ( i = 1, 2 ). For each ( y_i ), we solve ( f(x) = y_i ), which is a cubic equation. Each cubic can have one or three real roots.But we know that ( f(x_1) = x_1 ), which is one of the fixed points. So, if ( y_1 = x_1 ), then ( f(x) = y_1 ) has at least one real root at ( x = x_1 ). But since it's a cubic, it might have more.Wait, but ( y_1 ) and ( y_2 ) are solutions to ( 3y^2 + 2a y + b = 0 ). Is there a relationship between ( y_1 ) and ( y_2 ) and the fixed points of ( f(x) )?Given that ( f(x_1) = x_1 ), we can write:[ f(x_1) = x_1 ]Which implies:[ x_1^3 + a x_1^2 + b x_1 + c = x_1 ]Simplifying:[ x_1^3 + a x_1^2 + (b - 1) x_1 + c = 0 ]But I'm not sure how this helps directly.Maybe I should consider the fact that ( y_1 ) and ( y_2 ) are related to the critical values of ( f(x) ). Wait, the critical values are ( f(x_1) ) and ( f(x_2) ). Since ( f(x_1) = x_1 ), one of the critical values is ( x_1 ). The other critical value is ( f(x_2) ).So, the critical values are ( x_1 ) and ( f(x_2) ). Now, the equation ( 3y^2 + 2a y + b = 0 ) has solutions ( y_1 ) and ( y_2 ). Are these related to the critical values?Let me think about the relationship between ( y_1 ), ( y_2 ), ( x_1 ), and ( x_2 ). Since ( f'(x) = 3x^2 + 2a x + b ), and ( x_1 ) and ( x_2 ) are roots of ( f'(x) = 0 ), we can write:[ 3x_1^2 + 2a x_1 + b = 0 ][ 3x_2^2 + 2a x_2 + b = 0 ]So, both ( x_1 ) and ( x_2 ) satisfy the equation ( 3x^2 + 2a x + b = 0 ). But our quadratic equation in ( y ) is ( 3y^2 + 2a y + b = 0 ), which is the same form but with ( y ) instead of ( x ).This suggests that ( y_1 ) and ( y_2 ) are solutions to the same equation that ( x_1 ) and ( x_2 ) satisfy. Therefore, ( y_1 ) and ( y_2 ) are the same as ( x_1 ) and ( x_2 ), but possibly in a different order.Wait, that can't be right because ( y ) is ( f(x) ), not ( x ). So, ( y_1 ) and ( y_2 ) are values of ( f(x) ), not ( x ) itself. Therefore, ( y_1 ) and ( y_2 ) are not necessarily equal to ( x_1 ) and ( x_2 ), unless ( f(x) = x ) at those points, which is only true for ( x_1 ) as given.But we know that ( f(x_1) = x_1 ), so ( y_1 ) or ( y_2 ) could be equal to ( x_1 ). Let's check.Suppose ( y_1 = x_1 ). Then substituting into the quadratic equation:[ 3(x_1)^2 + 2a(x_1) + b = 0 ]But from the derivative, we have:[ 3(x_1)^2 + 2a(x_1) + b = 0 ]Which is exactly the same equation. So, ( y_1 = x_1 ) is indeed a solution to the quadratic equation. Therefore, one of the solutions ( y_1 ) or ( y_2 ) is equal to ( x_1 ).Similarly, if we consider ( y_2 = x_2 ), substituting into the quadratic equation:[ 3(x_2)^2 + 2a(x_2) + b = 0 ]Which is also true because ( x_2 ) is a root of the derivative. Therefore, the two solutions ( y_1 ) and ( y_2 ) are exactly ( x_1 ) and ( x_2 ).So, ( y_1 = x_1 ) and ( y_2 = x_2 ). Therefore, the equation ( 3(f(x))^2 + 2a f(x) + b = 0 ) simplifies to:[ 3(f(x))^2 + 2a f(x) + b = 0 ]Which has solutions ( f(x) = x_1 ) and ( f(x) = x_2 ).Now, for each of these, we need to find the number of real roots ( x ) such that ( f(x) = x_1 ) and ( f(x) = x_2 ).Starting with ( f(x) = x_1 ):[ x^3 + a x^2 + b x + c = x_1 ][ x^3 + a x^2 + b x + (c - x_1) = 0 ]We know that ( x = x_1 ) is a root because ( f(x_1) = x_1 ). So, ( x - x_1 ) is a factor of this cubic equation. Let's perform polynomial division or factorization to find the other roots.Let me denote the cubic as:[ (x - x_1)(x^2 + p x + q) = x^3 + a x^2 + b x + (c - x_1) ]Expanding the left side:[ x^3 + (p - x_1) x^2 + (q - p x_1) x - q x_1 ]Comparing coefficients with the right side:1. Coefficient of ( x^3 ): 1 = 1 (matches)2. Coefficient of ( x^2 ): ( p - x_1 = a ) ⇒ ( p = a + x_1 )3. Coefficient of ( x ): ( q - p x_1 = b ) ⇒ ( q = b + p x_1 = b + (a + x_1) x_1 = b + a x_1 + x_1^2 )4. Constant term: ( -q x_1 = c - x_1 ) ⇒ ( q = frac{c - x_1}{-x_1} = frac{x_1 - c}{x_1} )But from step 3, ( q = b + a x_1 + x_1^2 ). Therefore:[ b + a x_1 + x_1^2 = frac{x_1 - c}{x_1} ]Multiplying both sides by ( x_1 ):[ b x_1 + a x_1^2 + x_1^3 = x_1 - c ]Rearranging:[ x_1^3 + a x_1^2 + b x_1 + c = x_1 ]Which is exactly the equation ( f(x_1) = x_1 ), which is given. So, this consistency check works out.Therefore, the cubic equation ( f(x) = x_1 ) factors as:[ (x - x_1)(x^2 + (a + x_1) x + (b + a x_1 + x_1^2)) = 0 ]So, the other roots are the solutions to:[ x^2 + (a + x_1) x + (b + a x_1 + x_1^2) = 0 ]Let me denote this quadratic as:[ x^2 + (a + x_1) x + (b + a x_1 + x_1^2) = 0 ]The discriminant ( D_1 ) of this quadratic is:[ D_1 = (a + x_1)^2 - 4 times 1 times (b + a x_1 + x_1^2) ]Simplifying:[ D_1 = a^2 + 2a x_1 + x_1^2 - 4b - 4a x_1 - 4x_1^2 ][ D_1 = a^2 - 2a x_1 - 3x_1^2 - 4b ]Hmm, this seems complicated. Maybe there's a better way to analyze the number of real roots without computing the discriminant.Since ( f(x) = x_1 ) is a cubic equation with one known root ( x = x_1 ), the other two roots can be real or complex. The number of real roots depends on the behavior of the cubic function around ( x_1 ).Given that ( x_1 ) is a local maximum or minimum, the function ( f(x) ) will cross the line ( y = x_1 ) at ( x = x_1 ) and potentially at two other points. However, since ( x_1 ) is an extreme point, the function may only touch the line ( y = x_1 ) at that point, making it a double root, or it may cross it, giving three distinct roots.But wait, since ( f(x_1) = x_1 ), and ( x_1 ) is an extreme point, the function could be tangent to the line ( y = x ) at ( x = x_1 ), implying a double root. However, the problem doesn't specify whether ( x_1 ) is a maximum or minimum, so I can't be sure.Alternatively, since ( f(x) ) is a cubic, it must have at least one real root for ( f(x) = x_1 ), which is ( x = x_1 ). The other roots could be real or complex. But given that ( f(x) ) has two critical points, the function could have a local maximum above ( y = x_1 ) and a local minimum below ( y = x_1 ), or vice versa.Wait, no. Since ( x_1 ) is a critical point where ( f(x_1) = x_1 ), the function could be either increasing or decreasing around that point. If it's a local maximum, then to the left of ( x_1 ), the function is increasing, and to the right, it's decreasing. If it's a local minimum, the opposite.But regardless, the function ( f(x) ) will cross the line ( y = x_1 ) at least once at ( x = x_1 ). Depending on the shape of the cubic, it might cross again at two other points, giving three real roots, or it might only touch at ( x = x_1 ), giving a double root and one other real root.However, since ( f(x) ) is a cubic, it must have at least one real root for ( f(x) = x_1 ). But because ( x_1 ) is a critical point, it's possible that ( x = x_1 ) is a double root, meaning the cubic touches the line ( y = x_1 ) at ( x = x_1 ) and crosses it at another point. Therefore, in this case, ( f(x) = x_1 ) would have two real roots: one at ( x = x_1 ) (a double root) and another distinct root.But wait, a double root is still considered a single root with multiplicity two, but in terms of distinct real roots, it's just one. So, if ( f(x) = x_1 ) has a double root at ( x = x_1 ) and another distinct root, that would give two distinct real roots. However, if the cubic only crosses the line ( y = x_1 ) at ( x = x_1 ), then it would have only one real root.But I think in this case, since ( x_1 ) is a critical point and ( f(x_1) = x_1 ), the function could have a double root at ( x = x_1 ) and another real root. Therefore, ( f(x) = x_1 ) would have two distinct real roots: one at ( x = x_1 ) (with multiplicity two) and another at some ( x neq x_1 ).Wait, but multiplicity doesn't count towards distinctness. So, if ( x = x_1 ) is a double root, then the distinct real roots are ( x = x_1 ) and the other root, making it two distinct real roots. However, if the cubic only has ( x = x_1 ) as a simple root, then it would have three distinct real roots.This is getting a bit confusing. Maybe I should consider the graph of ( f(x) ) and the line ( y = x_1 ). Since ( x_1 ) is an extreme point, the function could be tangent to the line ( y = x_1 ) at ( x = x_1 ), meaning it's a double root, or it could cross the line, meaning it's a simple root.But without more information about the coefficients, it's hard to say for sure. However, since ( f(x) ) is a cubic, it must have at least one real root for ( f(x) = x_1 ), which is ( x = x_1 ). The other roots could be real or complex.But let's consider the other equation ( f(x) = x_2 ). Similarly, this is another cubic equation:[ x^3 + a x^2 + b x + c = x_2 ][ x^3 + a x^2 + b x + (c - x_2) = 0 ]This equation can have one or three real roots, depending on the value of ( x_2 ) relative to the critical points.But since ( x_2 ) is another critical point, and assuming ( x_1 ) is a local maximum and ( x_2 ) is a local minimum, the function ( f(x) ) will have a local maximum at ( x_1 ) and a local minimum at ( x_2 ). Therefore, the value ( f(x_2) ) is the local minimum value.So, if ( x_2 ) is the local minimum, then ( f(x_2) ) is less than or equal to the function values around it. Therefore, the equation ( f(x) = x_2 ) will have three real roots if ( x_2 ) is above the local minimum, but since ( x_2 ) is the local minimum, ( f(x) = x_2 ) will have at least one real root at ( x = x_2 ) and potentially two more.Wait, but ( x_2 ) is a critical point, so similar to ( x_1 ), the function could be tangent to the line ( y = x_2 ) at ( x = x_2 ), giving a double root, or it could cross it, giving three distinct roots.But again, without specific information, it's hard to determine. However, since ( f(x) ) is a cubic, it must have at least one real root for ( f(x) = x_2 ), which is ( x = x_2 ). The other roots could be real or complex.But let's think about the overall picture. We have two equations: ( f(x) = x_1 ) and ( f(x) = x_2 ). Each of these can have one or three real roots. However, since ( x_1 ) and ( x_2 ) are critical points, and ( f(x_1) = x_1 ), it's likely that ( f(x) = x_1 ) has a double root at ( x = x_1 ) and another real root, making it two distinct real roots. Similarly, ( f(x) = x_2 ) might have three distinct real roots if ( x_2 ) is above the local minimum, but since ( x_2 ) is the local minimum, it might only have one real root at ( x = x_2 ).Wait, no. If ( x_2 ) is the local minimum, then ( f(x_2) ) is the minimum value of the function. Therefore, the equation ( f(x) = x_2 ) will have exactly one real root at ( x = x_2 ) because the function is increasing on both sides of ( x_2 ) (since it's a local minimum). Therefore, ( f(x) = x_2 ) has only one real root.Similarly, for ( f(x) = x_1 ), since ( x_1 ) is a local maximum, the function is decreasing on both sides of ( x_1 ). Therefore, the equation ( f(x) = x_1 ) will have only one real root at ( x = x_1 ) because the function is decreasing on both sides, so it doesn't cross the line ( y = x_1 ) again.Wait, that contradicts what I thought earlier. Let me clarify.If ( x_1 ) is a local maximum, then to the left of ( x_1 ), the function is increasing, and to the right, it's decreasing. So, if ( f(x_1) = x_1 ), the function could cross the line ( y = x_1 ) at ( x = x_1 ) and potentially at two other points: one to the left and one to the right. But since ( x_1 ) is a local maximum, the function is increasing before ( x_1 ) and decreasing after ( x_1 ). Therefore, if ( f(x_1) = x_1 ), the function could cross the line ( y = x_1 ) at ( x = x_1 ) and potentially at two other points: one to the left (where the function was increasing towards ( x_1 )) and one to the right (where the function is decreasing away from ( x_1 )).However, since ( x_1 ) is a local maximum, the function value at ( x_1 ) is higher than the surrounding points. Therefore, if ( f(x_1) = x_1 ), the function could have a double root at ( x = x_1 ) if the function just touches the line ( y = x_1 ) there, or it could cross the line, giving three distinct roots.But without knowing the exact behavior, it's hard to say. However, since ( f(x) ) is a cubic, it must have at least one real root for ( f(x) = x_1 ), which is ( x = x_1 ). The other roots could be real or complex.But considering that ( f(x) ) has a local maximum at ( x_1 ), and ( f(x_1) = x_1 ), it's possible that the function only touches the line ( y = x_1 ) at ( x = x_1 ), making it a double root, and then crosses the line ( y = x_1 ) again at another point, giving a total of two distinct real roots.Similarly, for ( f(x) = x_2 ), since ( x_2 ) is a local minimum, the function could cross the line ( y = x_2 ) at ( x = x_2 ) and potentially at two other points. However, since ( x_2 ) is a local minimum, the function is decreasing before ( x_2 ) and increasing after ( x_2 ). Therefore, if ( f(x_2) = x_2 ), the function could have a double root at ( x = x_2 ) or cross the line ( y = x_2 ) at three distinct points.But again, without specific information, it's hard to determine. However, since ( f(x) ) is a cubic, it must have at least one real root for ( f(x) = x_2 ), which is ( x = x_2 ). The other roots could be real or complex.But let's consider the overall number of distinct real roots for the original equation ( 3(f(x))^2 + 2a f(x) + b = 0 ). This equation has two solutions for ( y = f(x) ): ( y = x_1 ) and ( y = x_2 ). For each of these, we solve ( f(x) = y ), which are cubic equations.If ( f(x) = x_1 ) has three distinct real roots and ( f(x) = x_2 ) has three distinct real roots, then the total number of distinct real roots for the original equation would be six. However, some of these roots might coincide, reducing the total number.But given that ( f(x) = x_1 ) and ( f(x) = x_2 ) are separate equations, their roots are distinct unless ( x_1 = x_2 ), which is not the case since ( x_1 ) and ( x_2 ) are distinct extreme points.Wait, but ( x_1 ) and ( x_2 ) are distinct, so ( f(x) = x_1 ) and ( f(x) = x_2 ) are separate equations with potentially separate roots. However, it's possible that some roots overlap if ( x_1 = x_2 ), but since they are distinct, this isn't the case.But earlier, I thought that ( f(x) = x_1 ) might have two distinct real roots (one at ( x = x_1 ) and another), and ( f(x) = x_2 ) might have one real root at ( x = x_2 ). So, in total, that would be three distinct real roots.Wait, no. If ( f(x) = x_1 ) has two distinct real roots (including ( x = x_1 )) and ( f(x) = x_2 ) has one real root at ( x = x_2 ), then the total number of distinct real roots would be three.But earlier, I considered that ( f(x) = x_1 ) could have three real roots, but given that ( x_1 ) is a local maximum, it's more likely to have two distinct real roots (one at ( x = x_1 ) and another). Similarly, ( f(x) = x_2 ) could have three real roots, but since ( x_2 ) is a local minimum, it might have one real root.But I'm getting conflicting thoughts here. Let me try a different approach.Since ( f(x) ) is a cubic, it can have up to three real roots for each equation ( f(x) = y ). However, the number of real roots depends on the value of ( y ) relative to the local maximum and minimum.Given that ( y_1 = x_1 ) and ( y_2 = x_2 ), and knowing that ( x_1 ) is a local maximum and ( x_2 ) is a local minimum, we can analyze the number of real roots for each ( f(x) = y_i ).For ( f(x) = x_1 ) (which is the local maximum value), the equation ( f(x) = x_1 ) will have exactly two real roots: one at ( x = x_1 ) (a double root) and another distinct root. This is because the function touches the line ( y = x_1 ) at ( x = x_1 ) and crosses it at another point.For ( f(x) = x_2 ) (which is the local minimum value), the equation ( f(x) = x_2 ) will have exactly two real roots: one at ( x = x_2 ) (a double root) and another distinct root. Similarly, the function touches the line ( y = x_2 ) at ( x = x_2 ) and crosses it at another point.Wait, but if both ( f(x) = x_1 ) and ( f(x) = x_2 ) have two real roots each, with one being a double root, then the total number of distinct real roots would be four: ( x = x_1 ), ( x = x_2 ), and two other distinct roots from each equation.But let me verify this. If ( f(x) = x_1 ) has a double root at ( x = x_1 ) and another root ( x = a ), and ( f(x) = x_2 ) has a double root at ( x = x_2 ) and another root ( x = b ), then the total distinct real roots are ( x_1 ), ( x_2 ), ( a ), and ( b ). So, four distinct real roots.But wait, could ( a ) and ( b ) coincide? That is, could the other roots from each equation be the same? If so, then the total number of distinct real roots would be three instead of four.But since ( x_1 ) and ( x_2 ) are distinct, and the function is a cubic, it's unlikely that the other roots ( a ) and ( b ) would coincide unless there's some symmetry, which isn't specified here.Therefore, it's more plausible that the total number of distinct real roots is four.Wait, but earlier I thought that ( f(x) = x_1 ) might have three real roots and ( f(x) = x_2 ) might have one, leading to a total of four. But now, considering that each equation has two real roots (one double and one simple), leading to four distinct roots.But I'm not entirely sure. Let me think about the graph again.Imagine the cubic function ( f(x) ) with a local maximum at ( x = x_1 ) where ( f(x_1) = x_1 ), and a local minimum at ( x = x_2 ) where ( f(x_2) ) is some value. The line ( y = x_1 ) will intersect the cubic at ( x = x_1 ) (a double root) and another point. Similarly, the line ( y = x_2 ) will intersect the cubic at ( x = x_2 ) (a double root) and another point.Therefore, each line intersects the cubic at two distinct points (one being a double root), leading to a total of four distinct intersection points.Hence, the original equation ( 3(f(x))^2 + 2a f(x) + b = 0 ) has four distinct real roots.But wait, the options given are A:3, B:4, C:5, D:6. So, according to this reasoning, the answer should be B:4.However, earlier I thought that each equation ( f(x) = y_i ) could have three real roots, leading to a total of six, but considering the double roots, it's four.But let me double-check. If ( f(x) = x_1 ) has a double root at ( x = x_1 ) and another root, and ( f(x) = x_2 ) has a double root at ( x = x_2 ) and another root, then the total number of distinct real roots is four: ( x_1 ), ( x_2 ), and the two other roots from each equation.Therefore, the correct answer should be B:4.But wait, the initial problem statement says "the number of distinct real roots of the equation ( 3(f(x))^2 + 2af(x) + b = 0 )". This equation is quadratic in ( f(x) ), so it has two solutions for ( f(x) ): ( y_1 = x_1 ) and ( y_2 = x_2 ). For each of these, solving ( f(x) = y_i ) gives up to three real roots, but considering the multiplicity and the fact that ( x_1 ) and ( x_2 ) are critical points, it's likely that each equation ( f(x) = y_i ) has two distinct real roots (one being a double root). Therefore, the total number of distinct real roots is four.But I'm still a bit unsure because I might be overcounting or undercounting the roots. Let me try to summarize:1. The equation ( 3(f(x))^2 + 2a f(x) + b = 0 ) has two solutions: ( f(x) = x_1 ) and ( f(x) = x_2 ).2. For ( f(x) = x_1 ): - ( x = x_1 ) is a root. - Depending on the function's behavior, there could be another root.3. For ( f(x) = x_2 ): - ( x = x_2 ) is a root. - Depending on the function's behavior, there could be another root.Given that ( x_1 ) and ( x_2 ) are distinct, and considering the function's shape, it's reasonable to assume that each equation ( f(x) = y_i ) contributes two distinct real roots, leading to a total of four distinct real roots.Therefore, the answer should be B:4.

💡Okay, so I have this problem where I'm given a function f(x) that's even, and for x greater than or equal to 0, f(x) is e^(2x). Then, I need to find the range of the real number k such that the function y = f(x) - |x - 1| - kx has exactly one root. Hmm, okay, let me try to unpack this step by step.First, since f(x) is even, that means f(-x) = f(x). So, for x less than 0, f(x) should be equal to f(-x), which is e^(2*(-x)) = e^(-2x). So, f(x) is e^(2x) when x is non-negative and e^(-2x) when x is negative. Got that.Now, the function y is given by f(x) - |x - 1| - kx. We need this function to have exactly one root, meaning it crosses the x-axis only once. So, I need to analyze y = f(x) - |x - 1| - kx and find the values of k for which this equation has only one solution.Since f(x) is defined differently for x >= 0 and x < 0, I think I should split the problem into two cases: when x is non-negative and when x is negative. Also, the absolute value function |x - 1| will behave differently depending on whether x is greater than or equal to 1 or less than 1. So, maybe I should further split the problem into three intervals: x < 0, 0 <= x < 1, and x >= 1.Let me start with x < 0. In this case, f(x) = e^(-2x). The absolute value |x - 1| becomes -(x - 1) because x - 1 is negative when x < 0. So, |x - 1| = -x + 1. Therefore, the function y becomes:y = e^(-2x) - (-x + 1) - kx = e^(-2x) + x - 1 - kx.Simplify that: y = e^(-2x) + (1 - k)x - 1.I need to find when this equation equals zero. So, e^(-2x) + (1 - k)x - 1 = 0.Hmm, this seems a bit complicated. Maybe I can analyze the behavior of this function as x approaches negative infinity and near x = 0.As x approaches negative infinity, e^(-2x) becomes e^(positive infinity), which goes to infinity. The term (1 - k)x - 1 will depend on the value of k. If k is less than 1, then (1 - k) is positive, so (1 - k)x - 1 will go to negative infinity. If k is greater than 1, then (1 - k) is negative, so (1 - k)x - 1 will go to positive infinity. If k = 1, then it's just -1.So, as x approaches negative infinity, if k < 1, y tends to infinity minus infinity, which is indeterminate. Hmm, maybe I should take the limit more carefully.Wait, e^(-2x) grows exponentially as x becomes more negative, while (1 - k)x - 1 is linear. So, regardless of k, e^(-2x) will dominate, and y will go to infinity as x approaches negative infinity.At x = 0, let's plug in x = 0: y = e^(0) + (1 - k)*0 - 1 = 1 - 1 = 0. So, y is zero at x = 0.So, the function y starts at infinity when x approaches negative infinity, comes down, and reaches zero at x = 0. Now, is there another root for x < 0? It depends on whether y crosses the x-axis somewhere else.To check that, maybe I can look at the derivative of y with respect to x for x < 0.The derivative dy/dx = derivative of e^(-2x) is -2e^(-2x), derivative of (1 - k)x is (1 - k), and derivative of -1 is 0. So, dy/dx = -2e^(-2x) + (1 - k).Now, since x < 0, -2x is positive, so e^(-2x) is greater than 1. Therefore, -2e^(-2x) is less than -2. So, dy/dx = -2e^(-2x) + (1 - k) is less than -2 + (1 - k) = (-1 - k).So, the derivative is negative if (-1 - k) is negative, which is always true because k is a real number. Wait, no, that's not necessarily true. Let me think again.Wait, dy/dx = -2e^(-2x) + (1 - k). Since e^(-2x) > 1 for x < 0, so -2e^(-2x) < -2. Therefore, dy/dx = something less than -2 + (1 - k). So, if (1 - k) is greater than 2, then dy/dx could be positive. Wait, but (1 - k) is a constant, so if 1 - k > 2, then k < -1. So, if k < -1, then 1 - k > 2, so dy/dx could be positive?Wait, let's plug in k < -1. Then, 1 - k is greater than 2, so dy/dx = -2e^(-2x) + (1 - k). Since e^(-2x) > 1, so -2e^(-2x) < -2. So, if (1 - k) > 2, then dy/dx could be positive if (1 - k) > 2e^(-2x). But e^(-2x) is greater than 1, so 2e^(-2x) is greater than 2. So, (1 - k) > 2e^(-2x) would require (1 - k) > something greater than 2. But (1 - k) is a constant, so unless (1 - k) is greater than the maximum of 2e^(-2x), which is infinity as x approaches negative infinity, which isn't possible. So, actually, dy/dx is always negative for x < 0.Wait, that can't be right. Let me think again.Wait, for x < 0, e^(-2x) is greater than 1, so -2e^(-2x) is less than -2. Then, (1 - k) is a constant. So, dy/dx = -2e^(-2x) + (1 - k). If (1 - k) is less than 2, then dy/dx is less than -2 + (1 - k), which is negative. If (1 - k) is greater than 2, then dy/dx could be positive or negative depending on x.Wait, but e^(-2x) is increasing as x becomes more negative, right? Because as x approaches negative infinity, -2x approaches positive infinity, so e^(-2x) approaches infinity. So, dy/dx = -2e^(-2x) + (1 - k). As x becomes more negative, -2e^(-2x) becomes more negative, so dy/dx becomes more negative. So, regardless of k, as x approaches negative infinity, dy/dx approaches negative infinity.But near x = 0, let's see. At x approaching 0 from the left, e^(-2x) approaches 1, so dy/dx approaches -2 + (1 - k) = (-1 - k). So, if (-1 - k) is negative, which is when k > -1, then dy/dx is negative near x = 0. If k < -1, then (-1 - k) is positive, so dy/dx is positive near x = 0.So, if k < -1, dy/dx is positive near x = 0, but as x becomes more negative, dy/dx becomes more negative. So, there must be a critical point somewhere where dy/dx = 0.Wait, so if k < -1, then dy/dx starts positive near x = 0, becomes negative as x decreases, so the function y has a maximum somewhere in x < 0. So, if the maximum of y is above zero, then y would cross the x-axis twice: once near x = 0 and once somewhere else. But we need y to have exactly one root. So, if k < -1, y might have two roots in x < 0, which is bad because we need only one root overall.But wait, at x = 0, y is zero. So, if y has another root in x < 0, that would be a second root, which we don't want. So, to have exactly one root, we need that y does not cross the x-axis again in x < 0. So, the maximum of y in x < 0 should be exactly at y = 0. That is, the function touches the x-axis at x = 0 and doesn't cross it again.So, to find when the maximum of y in x < 0 is zero, we can set the derivative dy/dx = 0 and y = 0 at the same point.Wait, but y is zero at x = 0, and if k < -1, y has a maximum somewhere in x < 0. So, to ensure that y doesn't cross the x-axis again, the maximum value of y in x < 0 should be zero. That is, the maximum occurs at x = 0, but y is zero there. So, actually, if the function is decreasing for all x < 0, then y would only cross the x-axis once at x = 0.But earlier, we saw that if k < -1, dy/dx is positive near x = 0, meaning y is increasing as x approaches 0 from the left. So, y would have a maximum at x = 0, but since y is zero there, it doesn't cross the x-axis again. Wait, but if y is increasing towards x = 0, then for x < 0, y would be less than zero? No, because at x approaching negative infinity, y approaches infinity. So, if y is increasing towards x = 0, starting from infinity, it would come down and reach zero at x = 0. So, in that case, y would cross the x-axis only once at x = 0.Wait, but if dy/dx is positive near x = 0, that means y is increasing as x increases towards 0. So, for x < 0, y is increasing towards zero. So, if y is increasing towards zero from below, then y would cross the x-axis only once at x = 0. But wait, at x approaching negative infinity, y approaches infinity, so y is coming down from infinity, increasing towards zero. So, if y is always increasing, it would cross the x-axis only once at x = 0.But earlier, I thought that if k < -1, dy/dx is positive near x = 0, but becomes negative as x decreases. So, y would have a maximum somewhere in x < 0. So, if y has a maximum above zero, it would cross the x-axis twice: once on the left of the maximum and once on the right. But since y approaches infinity as x approaches negative infinity, and y is zero at x = 0, if y has a maximum above zero, it would cross the x-axis once on the left of the maximum and once at x = 0. So, that would be two roots, which we don't want.Therefore, to have exactly one root, the maximum of y in x < 0 must be exactly zero. So, the maximum occurs at x = 0, where y is zero. That would happen if the derivative at x = 0 is zero, but y is zero there. Wait, but y is zero at x = 0 regardless.Wait, maybe I need to set the derivative at x = 0 to be zero to ensure that y doesn't have another root. So, dy/dx at x = 0 is -2e^(0) + (1 - k) = -2 + (1 - k) = -1 - k. If we set this equal to zero, then -1 - k = 0 => k = -1.So, when k = -1, the derivative at x = 0 is zero, meaning y has a horizontal tangent there. So, if k = -1, y is zero at x = 0, and the slope is zero there. So, does y cross the x-axis only once?Wait, let's check. If k = -1, then for x < 0, y = e^(-2x) + (1 - (-1))x - 1 = e^(-2x) + 2x - 1.At x = 0, y = 1 + 0 - 1 = 0.The derivative dy/dx = -2e^(-2x) + 2. At x = 0, dy/dx = -2 + 2 = 0.So, y has a critical point at x = 0. Let's see the behavior around x = 0.For x slightly less than 0, say x = -ε, where ε is a small positive number.y = e^(2ε) + 2*(-ε) - 1 ≈ (1 + 2ε) - 2ε - 1 = 0. So, near x = 0, y is approximately zero.But as x becomes more negative, y = e^(-2x) + 2x - 1. Since e^(-2x) grows exponentially, and 2x becomes more negative, but e^(-2x) dominates, so y tends to infinity as x approaches negative infinity.So, if k = -1, y is zero at x = 0, has a horizontal tangent there, and tends to infinity as x approaches negative infinity. So, does y cross the x-axis only once? It seems like y is zero at x = 0, and positive everywhere else for x < 0. So, y doesn't cross the x-axis again in x < 0, so the only root is at x = 0.But wait, let me check for x < 0, y = e^(-2x) + 2x - 1. Let's plug in x = -1: y = e^(2) + (-2) - 1 ≈ 7.389 - 3 ≈ 4.389 > 0.At x = -0.5: y = e^(1) + (-1) - 1 ≈ 2.718 - 2 ≈ 0.718 > 0.At x approaching 0 from the left: y approaches 0.So, y is positive for all x < 0 when k = -1, and zero at x = 0. So, y doesn't cross the x-axis again in x < 0, so the only root is at x = 0.Therefore, when k = -1, y has exactly one root at x = 0.But what if k > -1? Then, dy/dx at x = 0 is -1 - k. If k > -1, then -1 - k < 0. So, dy/dx is negative near x = 0, meaning y is decreasing as x approaches 0 from the left. So, y comes down from infinity, reaches zero at x = 0, and continues decreasing into negative values for x > 0.Wait, but for x > 0, we have a different expression for y. So, maybe I need to analyze x >= 0 as well.Okay, moving on to x >= 0. In this case, f(x) = e^(2x). The absolute value |x - 1| is equal to x - 1 if x >= 1, and 1 - x if 0 <= x < 1.So, we can split x >= 0 into two intervals: 0 <= x < 1 and x >= 1.First, let's consider 0 <= x < 1. Here, |x - 1| = 1 - x. So, y = e^(2x) - (1 - x) - kx = e^(2x) - 1 + x - kx = e^(2x) - 1 + (1 - k)x.So, y = e^(2x) + (1 - k)x - 1.We need to find when this equals zero: e^(2x) + (1 - k)x - 1 = 0.At x = 0: y = 1 + 0 - 1 = 0.So, y is zero at x = 0. Let's check the derivative here.dy/dx = 2e^(2x) + (1 - k).Since e^(2x) is always positive, and 2e^(2x) is always greater than 2 for x > 0. So, dy/dx = 2e^(2x) + (1 - k). If (1 - k) is greater than -2, then dy/dx is positive. If (1 - k) is less than -2, then dy/dx could be positive or negative.Wait, for x >= 0, e^(2x) >= 1, so 2e^(2x) >= 2. Therefore, dy/dx = 2e^(2x) + (1 - k) >= 2 + (1 - k) = 3 - k.So, if 3 - k > 0, which is when k < 3, then dy/dx is positive. If k >= 3, then dy/dx could be non-positive.Wait, but 2e^(2x) is always greater than 2, so even if k is large, 2e^(2x) + (1 - k) could still be positive.Wait, let's think about it. For x >= 0, 2e^(2x) is increasing, so the minimum value of dy/dx occurs at x = 0, which is 2e^0 + (1 - k) = 2 + (1 - k) = 3 - k.So, if 3 - k > 0, i.e., k < 3, then dy/dx is always positive for x >= 0. If k = 3, then dy/dx at x = 0 is zero, and for x > 0, dy/dx is positive because 2e^(2x) > 2. If k > 3, then dy/dx at x = 0 is negative, but for x > 0, dy/dx becomes positive because 2e^(2x) grows exponentially.So, let's analyze the behavior of y in 0 <= x < 1.At x = 0, y = 0.If k < 3, dy/dx is positive, so y is increasing. Therefore, y starts at zero and increases as x increases. So, in 0 <= x < 1, y is zero at x = 0 and increases, so it doesn't cross the x-axis again in this interval.If k = 3, dy/dx at x = 0 is zero, but for x > 0, dy/dx becomes positive. So, y has a horizontal tangent at x = 0, but then increases. So, y is zero at x = 0 and increases, so no other roots in this interval.If k > 3, dy/dx at x = 0 is negative, so y starts decreasing from zero. But since dy/dx becomes positive for x > 0, y will have a minimum somewhere in 0 < x < 1. If the minimum is below zero, y will cross the x-axis twice: once at x = 0 and once somewhere else in 0 < x < 1. If the minimum is exactly zero, y will touch the x-axis at the minimum point, giving a repeated root. If the minimum is above zero, y will not cross the x-axis again.But we need y to have exactly one root overall. So, in the interval 0 <= x < 1, if k > 3, y could have another root, which we don't want. So, to prevent that, we need that y doesn't cross the x-axis again in 0 <= x < 1. That would happen if the minimum of y in this interval is above zero.Wait, but y is zero at x = 0, and if k > 3, y starts decreasing, reaches a minimum, and then increases. So, if the minimum is below zero, y crosses the x-axis twice. If the minimum is exactly zero, it's a repeated root, which still counts as one root. If the minimum is above zero, y doesn't cross the x-axis again.Wait, but y is zero at x = 0, so if the minimum is above zero, y would be positive for all x > 0, which would mean y only crosses the x-axis once at x = 0. But if the minimum is below zero, y crosses the x-axis twice: once at x = 0 and once somewhere else.But wait, y is zero at x = 0, and if k > 3, y starts decreasing, so it goes below zero, reaches a minimum, and then increases. So, it must cross the x-axis again somewhere in 0 < x < 1. Therefore, for k > 3, y has two roots in x >= 0: one at x = 0 and another in 0 < x < 1. But we need exactly one root overall, so we must have k <= 3.Wait, but when k = 3, y has a horizontal tangent at x = 0, and then increases. So, y is zero at x = 0 and positive elsewhere, so only one root.If k < 3, y is increasing for x >= 0, so y is zero at x = 0 and positive elsewhere, so only one root.If k > 3, y starts decreasing, goes below zero, then increases, crossing the x-axis again, leading to two roots in x >= 0, which we don't want.Therefore, for x >= 0, to have exactly one root, k must be <= 3.Now, let's consider x >= 1. Here, |x - 1| = x - 1, so y = e^(2x) - (x - 1) - kx = e^(2x) - x + 1 - kx = e^(2x) - (1 + k)x + 1.So, y = e^(2x) - (1 + k)x + 1.We need to find when this equals zero: e^(2x) - (1 + k)x + 1 = 0.Let's analyze the behavior of this function.As x approaches infinity, e^(2x) dominates, so y approaches infinity. At x = 1, y = e^2 - (1 + k)*1 + 1 = e^2 - k.So, y at x = 1 is e^2 - k.Now, let's check the derivative dy/dx = 2e^(2x) - (1 + k).Since e^(2x) is increasing, dy/dx is increasing. At x = 1, dy/dx = 2e^2 - (1 + k).If 2e^2 - (1 + k) > 0, which is always true because 2e^2 is about 14.778, so 14.778 - (1 + k) > 0 implies k < 13.778, which is always true since k is a real number.Wait, but more importantly, the derivative dy/dx = 2e^(2x) - (1 + k). Since 2e^(2x) is always greater than 2 for x > 0, so if (1 + k) < 2e^(2x), which is always true for x > 0 because 2e^(2x) grows exponentially. So, dy/dx is always positive for x > 0, meaning y is strictly increasing for x > 0.Therefore, y = e^(2x) - (1 + k)x + 1 is strictly increasing for x >= 1. So, it can have at most one root in x >= 1.Now, let's see if y ever crosses zero in x >= 1.At x = 1, y = e^2 - k.If e^2 - k > 0, which is when k < e^2 (~7.389), then y is positive at x = 1 and increasing, so it doesn't cross zero in x >= 1.If e^2 - k = 0, which is when k = e^2, then y is zero at x = 1 and positive elsewhere, so it's a root at x = 1.If e^2 - k < 0, which is when k > e^2, then y is negative at x = 1, but since y approaches infinity as x approaches infinity, it must cross zero somewhere in x > 1.But wait, since y is strictly increasing, if y is negative at x = 1 and positive as x approaches infinity, it must cross zero exactly once in x > 1.So, for x >= 1, y has exactly one root if k > e^2, and no roots if k <= e^2.But we need the entire function y = f(x) - |x - 1| - kx to have exactly one root. So, we have to consider the roots in all intervals.From x < 0: If k >= -1, y has exactly one root at x = 0. If k < -1, y has two roots: one at x = 0 and another in x < 0.From x >= 0: If k <= 3, y has exactly one root at x = 0. If k > 3, y has two roots: one at x = 0 and another in 0 < x < 1.From x >= 1: If k > e^2, y has one root in x > 1. If k <= e^2, y has no roots in x >= 1.So, combining all these:- If k < -1: y has two roots (x < 0 and x = 0) and possibly another in x >= 0 or x >= 1. Wait, no, if k < -1, y has two roots in x < 0 and x = 0, but in x >= 0, y has one root at x = 0 and possibly another in 0 < x < 1 if k > 3. So, total roots could be more than one.Wait, this is getting complicated. Let me try to summarize:We need y to have exactly one root overall. So, we need to ensure that in each interval, y doesn't cross the x-axis more than once, and that the total number of roots is one.From x < 0:- If k >= -1: y has exactly one root at x = 0.- If k < -1: y has two roots (x < 0 and x = 0).From x >= 0:- If k <= 3: y has exactly one root at x = 0.- If k > 3: y has two roots (x = 0 and another in 0 < x < 1).From x >= 1:- If k > e^2: y has one root in x > 1.- If k <= e^2: y has no roots in x >= 1.So, to have exactly one root overall, we need:1. From x < 0: k >= -1 (so only one root at x = 0).2. From x >= 0: k <= 3 (so only one root at x = 0).3. From x >= 1: k <= e^2 (so no roots in x >= 1).But wait, if k <= 3, then from x >= 1, y has no roots if k <= e^2, which is always true since 3 < e^2 (~7.389). So, if k <= 3, y has no roots in x >= 1.But if k > 3, y has another root in 0 < x < 1, which we don't want.Also, if k < -1, y has another root in x < 0, which we don't want.Therefore, to have exactly one root overall, we need:- k >= -1 (to prevent another root in x < 0).- k <= 3 (to prevent another root in 0 < x < 1).- k <= e^2 (to prevent another root in x >= 1). But since 3 < e^2, the stricter condition is k <= 3.Therefore, combining these, the range of k is -1 <= k <= 3.Wait, but let me double-check.If k is between -1 and 3, inclusive, then:- For x < 0: y has exactly one root at x = 0.- For x >= 0: y has exactly one root at x = 0.- For x >= 1: y has no roots.So, total roots: one.If k < -1:- For x < 0: y has two roots.- For x >= 0: y has one root.Total roots: three.If k > 3:- For x < 0: y has one root.- For x >= 0: y has two roots.- For x >= 1: y has one root.Total roots: four.Wait, that can't be right. Wait, no, for k > 3:- x < 0: y has one root at x = 0.- x >= 0: y has two roots (x = 0 and another in 0 < x < 1).- x >= 1: y has one root.So, total roots: three.But we need exactly one root, so k must be between -1 and 3.Wait, but when k = -1:- x < 0: y has one root at x = 0.- x >= 0: y has one root at x = 0.- x >= 1: y has no roots.So, total roots: one.Similarly, when k = 3:- x < 0: y has one root at x = 0.- x >= 0: y has one root at x = 0.- x >= 1: y has no roots.Total roots: one.Therefore, the range of k is -1 <= k <= 3.So, the final answer is k ∈ [-1, 3].

💡First, I need to find the intersection of sets ( A ) and ( B ), denoted as ( A cap B ). This involves identifying the elements that are common to both sets.Next, I need to determine the complement of set ( B ) with respect to the universal set ( U ), written as ( complement_U B ). This means finding all the elements in ( U ) that are not in ( B ).Finally, I will find the union of set ( A ) and the complement of ( B ), which is ( A cup (complement_U B) ). This involves combining all the elements from both sets, ensuring there are no duplicate elements in the final result.

💡Alright, so I've got this function here: f(x) = sin(x + 7π/4) + cos(x - 3π/4). I need to figure out its period and whether its graph is symmetric about a specific line. The options are A, B, C, D, each giving a period and a line of symmetry.First, I remember that both sine and cosine functions have a period of 2π, but sometimes when you combine them, the period can be smaller. So maybe I need to see if this function simplifies into something with a smaller period.Let me try to expand these sine and cosine terms using the angle addition formulas. For sin(x + 7π/4), I can use sin(A + B) = sinA cosB + cosA sinB. Similarly, for cos(x - 3π/4), I can use cos(A - B) = cosA cosB + sinA sinB.So, expanding sin(x + 7π/4):sin(x)cos(7π/4) + cos(x)sin(7π/4).And expanding cos(x - 3π/4):cos(x)cos(3π/4) + sin(x)sin(3π/4).Now, I need to compute these cosine and sine values. Let's recall that 7π/4 is in the fourth quadrant, so cosine is positive and sine is negative there. Similarly, 3π/4 is in the second quadrant, so cosine is negative and sine is positive.Calculating each term:cos(7π/4) = √2/2,sin(7π/4) = -√2/2,cos(3π/4) = -√2/2,sin(3π/4) = √2/2.Substituting these back into the expanded expressions:sin(x + 7π/4) = sin(x)(√2/2) + cos(x)(-√2/2) = (√2/2)sin(x) - (√2/2)cos(x).cos(x - 3π/4) = cos(x)(-√2/2) + sin(x)(√2/2) = (-√2/2)cos(x) + (√2/2)sin(x).Now, adding these two results together:f(x) = [ (√2/2)sin(x) - (√2/2)cos(x) ] + [ (-√2/2)cos(x) + (√2/2)sin(x) ].Combine like terms:sin(x) terms: (√2/2 + √2/2)sin(x) = √2 sin(x).cos(x) terms: (-√2/2 - √2/2)cos(x) = -√2 cos(x).So, f(x) simplifies to √2 sin(x) - √2 cos(x). I can factor out √2:f(x) = √2 [sin(x) - cos(x)].Hmm, maybe I can write this as a single sine function with a phase shift. I remember that expressions like A sin(x) + B cos(x) can be written as C sin(x + φ), where C = √(A² + B²) and φ is the phase shift.In this case, A = 1 and B = -1, so C = √(1 + 1) = √2. The phase shift φ can be found using tanφ = B/A, which is -1/1 = -1. So φ = -π/4.Therefore, f(x) = √2 sin(x - π/4).Now, the period of sin(x - π/4) is 2π, so the period of f(x) is also 2π. That means options A and C, which say the period is π, are incorrect. So we're down to B and D.Next, I need to check the symmetry. The function is f(x) = √2 sin(x - π/4). The graph of a sine function is symmetric about its midline, which in this case is the x-axis shifted by the phase. But the question is about symmetry about a vertical line, not the x-axis.For a sine function, the graph is symmetric about the line x = kπ + π/2, where k is an integer. Wait, no, actually, the sine function is symmetric about its peaks and troughs. For sin(x - π/4), the maximum occurs at x = π/4 + π/2 = 3π/4, and the minimum at x = π/4 + 3π/2 = 7π/4. So the lines of symmetry would be halfway between these points.But maybe a better approach is to consider the function f(x) = √2 sin(x - π/4). For a function to be symmetric about a vertical line x = a, it must satisfy f(a + h) = f(a - h) for all h.Let me test this. Let's suppose the line of symmetry is x = a. Then:√2 sin((a + h) - π/4) = √2 sin((a - h) - π/4).Divide both sides by √2:sin(a + h - π/4) = sin(a - h - π/4).Using the identity sin(θ) = sin(π - θ), so:sin(a + h - π/4) = sin(π - (a - h - π/4)).Simplify the right side:sin(π - a + h + π/4) = sin((5π/4 - a) + h).So, for the equality sin(a + h - π/4) = sin((5π/4 - a) + h) to hold for all h, the arguments must differ by an integer multiple of 2π or be supplementary angles.But since this must hold for all h, the only way is if a + h - π/4 = (5π/4 - a) + h + 2πk, which simplifies to:a - π/4 = 5π/4 - a + 2πk.Solving for a:2a = 6π/4 + 2πk = 3π/2 + 2πk.Thus, a = 3π/4 + πk.So the lines of symmetry are at x = 3π/4 + πk, where k is any integer.Looking back at the options, D says the graph is symmetric about x = -π/4. Let's check if -π/4 is equal to 3π/4 + πk for some integer k.Set 3π/4 + πk = -π/4:πk = -π/4 - 3π/4 = -π.So k = -1.Yes, when k = -1, a = 3π/4 - π = -π/4.Therefore, the graph is symmetric about x = -π/4, which is option D.So, putting it all together, the period is 2π and the graph is symmetric about x = -π/4. Therefore, the correct answer is D.

💡First, I need to find the perfect-square integer closest to 325. To do this, I'll start by identifying the integers whose squares are near 325.I know that the square of 18 is 324, which is just one less than 325. The square of 19 is 361, which is significantly larger than 325. Comparing these two squares to 325, 324 is much closer than 361.Therefore, the perfect-square integer closest to 325 is 324.